Find the mean and variance of the sum Y=∑i=1^5 Xi, where X1, …, X5 are iid, having pdf f(x)=6 x(

Find the mean and variance of the sum Y=∑i=1^5 Xi, where X1,...

Key Concepts

Variance of Independent Random Variables

When random variables are independent, the variance of their sum is equal to the sum of their individual variances. This property is critical in assessing spread or dispersion in the sum of outcomes, and it relies on the fact that there are no correlation (covariance) terms to account for among independent variables.

Beta Distribution

The Beta distribution is a family of continuous probability distributions defined on the interval (0,1) and characterized by two shape parameters. In this context, the given density function corresponds to a Beta distribution, and knowing its parameters allows one to directly compute the mean and variance using well-known formulas. This is particularly useful in Bayesian statistics and modeling proportions or probabilities.

Independence and Sum of Random Variables

In probability theory, when random variables are independent and identically distributed (iid), the mean of their sum is the sum of their individual means and the variance of their sum is the sum of their individual variances. This principle greatly simplifies the computation of moments for sums of random variables, as it allows each moment to be calculated without considering covariance terms.

Linearity of Expectation

The concept of linearity of expectation states that the expected value of a sum of random variables is equal to the sum of the expected values of each individual variable, regardless of any dependence between them. This is a foundational concept in probability and is widely used in many areas including statistics and finance for the simplification of computations of means.

Transcript

00:01 Hi, here it is given that x1, x2 up to x5, there are five random variables, their iid following the pdf it is given by and we have to find the mean and variance of the sum that is y equals to summation irons from 1 to 5.

00:21 I write it down, y equals to summation irons from 1 to 5 x.

00:26 We have to find the mean and variance of this random variable y, okay, where x1, x2, x2, up to x5, these are iid follows a probability density function f x where f x is given by is equals to 6x into 1 minus x where 0 less than x less than 1 less than 1.

00:58 And 0 otherwise.

01:02 Okay.

01:03 Now, clearly, see, fx can be written as x to the power 2 minus 1 into 1 minus x to 3 power 2 minus 1, whole divided by beta 2 comma 2, right? 0 less than x less than 1.

01:22 Where beta, we know, just a minute, we know beta mn is equals to gamma n is equal to gamma m into gamma n whole divided by gamma m plus n where gamma n is equal to n minus 1 factorial now clearly if we put both the values of m and n is equal to 2 we get see beta 2 comma 2 this quantity is equals to gamma 2, gamma 2 whole divided by gamma 4.

02:04 Now gamma 2 means 1 factorial into 1 factorial 1 factorial 1 divided by this is gamma 4 means 3 factorial 3 factorial is 6.

02:12 So 1 by 6.

02:13 See if i put 1 by 6 here 6 will be in the numerator.

02:17 So this fx is same as the overall density function of beta distribution of first kind with parameters 2 and 2.

02:28 So, clearly the random variable each follows beta 1 distribution with parameters 2 comma.

02:38 Now, see, we have to find the mean and variance of this random variable y.

02:45 So, expectation of y and variance of y we have to find.

02:49 So, see, expectation of y, this quantity is equal to expectation of summation irons from 1 to 5 x i right so this is summation irons from 1 to 5 expectation of x i similarly variants of y this is equals to variance of summation irons from 1 to 5 xi so this is equals to variance will go inside and the covariance part will be 0 so it will be only summation variance x i this is equals to we have to find the value of variance x i to find the variance of y okay now see expectation of x is equals to integration it will be the support of x is 0 to 1 x into f x d x so it will be integration 0 to 1 x into f x x x 6 x 1 minus x d x x 6 is a constant so it comes out to integration 0 to 1, x square 1 minus x d x.

04:14 Now see clearly i'll write down the beta 1 distribution's pdf.

04:22 See we can write beta mn is equal to integration 0 to 1 x to the power m minus 1 d x...

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